Prove theoretically the relation between emf induced in a coil and rate of Change of magnetic flux in electromagnetic

Let us consider a rectangular wire loop PQRS of width $l$, and its plane perpendicular to a uniform magnetic field. The loop is being pulled out of magnetic field at a constant speed $v$. At any instant, let '$x$' be the length of loop in the magnetic field. As the loop moves towards right, the area of the loop inside the field changes by $\mathrm{dA}=\mathrm{ldx}=\mathrm{lvdt}$.

So the change in magnetic flux is given by
$$\begin{gathered} \mathrm{d\varphi }=\mathrm{BdA}=\mathrm{Blvdt} \\ \mathrm{so}, \frac{\mathrm{d\varphi }}{\mathrm{dt}}=\frac{\mathrm{Blvdt}}{\mathrm{dt}}=\mathrm{Bvl} \end{gathered}$$
 

Now a straight current-carrying conductor of length $L$ experiences a magnetic force given by
$\overrightarrow{\mathrm{F}}=\mathrm{I}\overrightarrow{\mathrm{L}}\times \overrightarrow{\mathrm{B}}$
Its direction is given by Fleming's left and rule.
So, the forces F₁​ and F₂ on wires $PR$ and $QS$ respectively are equal in magnitude and opposite in direction and have the same line of action. Hence they balance each other. There is no force on the wire $RS$ as it lies outside the field. The force F₃​ on the wire $PQ$ has magnitude F₃ = IBL directed towards left. To move the loop with constant velocity $v$an external force must be applied. So work done by the external agent is given by
$$\begin{gathered} \mathrm{dw}=\mathrm{Fdx}=-\mathrm{IlBdx}=-\mathrm{IBdA} \\ \mathrm{dw}=-{\mathrm{Id\varphi }}_{\mathrm{m}} \\ \mathrm{So} \mathrm{rate} \mathrm{of} \mathrm{doing} \mathrm{work}, \mathrm{P}=\frac{\mathrm{dw}}{\mathrm{dt}} \\ \mathrm{P}=\mathrm{I}\left(\frac{-{\mathrm{d\varphi }}_{\mathrm{m}}}{\mathrm{dt}}\right) \\ \mathrm{But}, \mathrm{P}=\mathrm{EI} \\ \mathrm{so} \mathrm{E}=-\frac{{\mathrm{d\varphi }}_{\mathrm{m}}}{\mathrm{dt}} \end{gathered}$$

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